Solve 35-64n=-21n^2

$$35 - 64 n = - 21 n ^ { 2 }$$

Solve for n

$$n=\frac{5}{7}\approx 0.714285714$$

$$n = \frac{7}{3} = 2\frac{1}{3} \approx 2.333333333$$

Steps Using Factoring By Grouping

Add 21n^{2} to both sides.

$$35-64n+21n^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$21n^{2}-64n+35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 21n^{2}+an+bn+35. To find a and b, set up a system to be solved.

$$a+b=-64 ab=21\times 35=735$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 735.

$$-1,-735 -3,-245 -5,-147 -7,-105 -15,-49 -21,-35$$

Calculate the sum for each pair.

$$-1-735=-736 -3-245=-248 -5-147=-152 -7-105=-112 -15-49=-64 -21-35=-56$$

The solution is the pair that gives sum -64.

$$a=-49 b=-15$$

Rewrite 21n^{2}-64n+35 as \left(21n^{2}-49n\right)+\left(-15n+35\right).

$$\left(21n^{2}-49n\right)+\left(-15n+35\right)$$

Factor out 7n in the first and -5 in the second group.

$$7n\left(3n-7\right)-5\left(3n-7\right)$$

Factor out common term 3n-7 by using distributive property.

$$\left(3n-7\right)\left(7n-5\right)$$

To find equation solutions, solve 3n-7=0 and 7n-5=0.

$$n=\frac{7}{3} n=\frac{5}{7}$$

Steps Using the Quadratic Formula

Add 21n^{2} to both sides.

$$35-64n+21n^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$21n^{2}-64n+35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 21 for a, -64 for b, and 35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-64\right)±\sqrt{\left(-64\right)^{2}-4\times 21\times 35}}{2\times 21}$$

Square -64.

$$n=\frac{-\left(-64\right)±\sqrt{4096-4\times 21\times 35}}{2\times 21}$$

Multiply -4 times 21.

$$n=\frac{-\left(-64\right)±\sqrt{4096-84\times 35}}{2\times 21}$$

Multiply -84 times 35.

$$n=\frac{-\left(-64\right)±\sqrt{4096-2940}}{2\times 21}$$

Add 4096 to -2940.

$$n=\frac{-\left(-64\right)±\sqrt{1156}}{2\times 21}$$

Take the square root of 1156.

$$n=\frac{-\left(-64\right)±34}{2\times 21}$$

The opposite of -64 is 64.

$$n=\frac{64±34}{2\times 21}$$

Multiply 2 times 21.

$$n=\frac{64±34}{42}$$

Now solve the equation n=\frac{64±34}{42} when ± is plus. Add 64 to 34.

$$n=\frac{98}{42}$$

Reduce the fraction \frac{98}{42} to lowest terms by extracting and canceling out 14.

$$n=\frac{7}{3}$$

Now solve the equation n=\frac{64±34}{42} when ± is minus. Subtract 34 from 64.

$$n=\frac{30}{42}$$

Reduce the fraction \frac{30}{42} to lowest terms by extracting and canceling out 6.

$$n=\frac{5}{7}$$

The equation is now solved.

$$n=\frac{7}{3} n=\frac{5}{7}$$

Steps for Completing the Square

Add 21n^{2} to both sides.

$$35-64n+21n^{2}=0$$

Subtract 35 from both sides. Anything subtracted from zero gives its negation.

$$-64n+21n^{2}=-35$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$21n^{2}-64n=-35$$

Divide both sides by 21.

$$\frac{21n^{2}-64n}{21}=-\frac{35}{21}$$

Dividing by 21 undoes the multiplication by 21.

$$n^{2}-\frac{64}{21}n=-\frac{35}{21}$$

Reduce the fraction \frac{-35}{21} to lowest terms by extracting and canceling out 7.

$$n^{2}-\frac{64}{21}n=-\frac{5}{3}$$

Divide -\frac{64}{21}, the coefficient of the x term, by 2 to get -\frac{32}{21}. Then add the square of -\frac{32}{21} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-\frac{64}{21}n+\left(-\frac{32}{21}\right)^{2}=-\frac{5}{3}+\left(-\frac{32}{21}\right)^{2}$$

Square -\frac{32}{21} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{64}{21}n+\frac{1024}{441}=-\frac{5}{3}+\frac{1024}{441}$$

Add -\frac{5}{3} to \frac{1024}{441} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}-\frac{64}{21}n+\frac{1024}{441}=\frac{289}{441}$$

Factor n^{2}-\frac{64}{21}n+\frac{1024}{441}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{32}{21}\right)^{2}=\frac{289}{441}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{32}{21}\right)^{2}}=\sqrt{\frac{289}{441}}$$

Simplify.

$$n-\frac{32}{21}=\frac{17}{21} n-\frac{32}{21}=-\frac{17}{21}$$

Add \frac{32}{21} to both sides of the equation.

$$n=\frac{7}{3} n=\frac{5}{7}$$